LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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One-step estimate for projected subgradient descent

Statement

Under the projected-subgradient update of Projected subgradient method with finite budget, let $x^{*} \in C$ and suppose the $k$ th oracle call returns $g_{k} \in \partial f (x_{k})$ . Then $∥ x_{k + 1} - x^{*} ∥_{2}^{2} \leq ∥ x_{k} - x^{*} ∥_{2}^{2} - 2 α_{k} (f (x_{k}) - f (x^{*})) + α_{k}^{2} ∥ g_{k} ∥_{2}^{2} .$

Facts & Assumptions

Given: The projected update Projected subgradient method with finite budget, subgradient convention Subgradients and the subdifferential of a convex function, and projection variational inequality Metric projection onto a closed convex set satisfies the variational inequality.

[F1]

For $p = P_{C} (u)$ and every $z \in C$ , $⟨ u - p, z - p ⟩ \leq 0$ (Metric projection onto a closed convex set satisfies the variational inequality).

[F2]

$g_{k} \in \partial f (x_{k})$ implies $f (x^{*}) \geq f (x_{k}) + ⟨ g_{k}, x^{*} - x_{k} ⟩$ (Subgradients and the subdifferential of a convex function).

Proof

1.1F1givenalgebra

Put $u = x_{k} - α_{k} g_{k}$ and $p = x_{k + 1} = P_{C} (u)$ . Applying [F1] with $z = x^{*}$ gives $⟨ u - p, x^{*} - p ⟩ \leq 0$ , hence $⟨ u - p, p - x^{*} ⟩ \geq 0$ . Expanding $∥ u - x^{*} ∥_{2}^{2} = ∥ u - p ∥_{2}^{2} + ∥ p - x^{*} ∥_{2}^{2} + 2 ⟨ u - p, p - x^{*} ⟩$ yields $∥ p - x^{*} ∥_{2}^{2} \leq ∥ u - x^{*} ∥_{2}^{2}$ .

2.1F2step 1.1algebra∎

Expand the right side as $∥ x_{k} - x^{*} ∥_{2}^{2} - 2 α_{k} ⟨ g_{k}, x_{k} - x^{*} ⟩ + α_{k}^{2} ∥ g_{k} ∥_{2}^{2}$ . By [F2], $⟨ g_{k}, x_{k} - x^{*} ⟩ \geq f (x_{k}) - f (x^{*})$ ; substituting proves the estimate.

Depends on

Used by

Projected subgradient best-iterate and weighted-average bound Theorem

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sébastien Bubeck, Convex Optimization: Algorithms and Complexity, §3.1 (standard reference, not scraped)
Dimitri Bertsekas, MIT 6.253 Complete Lecture Notes, Lecture 15 (standard reference, not scraped)